HDU 1016（搜索题，DFS）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5720    Accepted Submission(s): 2553

Problem Description

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.

 

 

Input

n (0 < n < 20).

 

 

Output

The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.

You are to write a program that completes above process.

Print a blank line after each case.

 

 

Sample Input

   
   
   
   

    
    
    
    6
8
   
   
   
   

 

 

Sample Output

   
   
   
   

    
    
    
    Case 1:
1 4 3 2 5 6
1 6 5 2 3 4

Case 2:
1 2 3 8 5 6 7 4
1 2 5 8 3 4 7 6
1 4 7 6 5 8 3 2
1 6 7 4 3 8 5 2
   
   
   
   
   
   
   
   

    
    
    
    此题纠结了我一个晚上。。。太悲剧了~ 一道简单的DFS题而已！
   
   
   
   
   
   
   
   

    
    
    #include <iostream>
#include <math.h>
using namespace std;
#define N 20

int map[N][N],mark[N];
int res[N],p[N];
int n,ct;

bool is_prime(int u)
{
	if(u==0 || u==1) return false;
	if(u==2) return true;
	if(u%2==0) return false;
	for(int i=3;i<=sqrt((double)u);i+=2)
		if(u%i==0) 
			return false;
		return true;
}

void init()
{
	int i,j;
	for(i=1;i<N;i++)
	{
		for(j=i+1;j<N;j++)
		{
			if(is_prime(i+j))
				map[i][j]=map[j][i]=1;
		}
	}
}

void output(int t)//输出例程
{
    int k=0;
    while(t)//通过对路径的分析获得结果
    {
        res[k++]=t;
        t=p[t];
    }
    for(int j=n-1;j>=0;j--)//输出出来
    {
        if(j!=n-1)
            putchar(' ');
        printf("%d",res[j]);
    }
    putchar('/n');
    
}


void DFS(int x,int count)
{
	mark[x]=1;
	if(count==n && map[x][1]==1)
	{
		output(x);
	}
	for (int i=1;i<=n;i++)
	{
		if(map[x][i]==1 && mark[i]==0)
		{
			p[i]=x;
			DFS(i,count+1);
			mark[i]=0;
		}
	}
}

int main()
{
	init();
	while (scanf("%d",&n)!=EOF)
	{
		ct++;
		printf("Case %d:/n",ct);
		DFS(1,1);
		printf("/n");
	}
	return 0;
}